Last updated at Dec. 27, 2021 by

Transcript

Ex 9.4, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : ππ¦/ππ₯=(1 β cosβ‘π₯)/(1 + cosβ‘π₯ ) ππ¦/ππ₯ = (1 β cosβ‘π₯)/(1 + cosβ‘π₯ ) We know that cos 2x = 2cos2 x β 1 Putting x = π₯/2 cos 2π₯/2 = 2 cos2 π₯/2 β 1 cos x = 2 cos2 π₯/2 β 1 1 + cos x = 2cos2 π₯/2 We know cos 2x = 1 β 2sin2 x Putting x = π₯/2 cos2 2π₯/2 = 1 β 2 sin2 π₯/2 cos x = 1 β 2sin2 π₯/2 1 β cos x = 2sin2 π₯/2 Putting values in equation π π/π π = (π γγπ¬π’π§γ^π γβ‘γπ/πγ )/(π γγππ¨π¬γ^π γβ‘γπ/πγ ) ππ¦ = (γsin^2 γβ‘γπ₯/2γ )/cos^(2 )β‘γπ₯/2γ ππ₯ ππ¦ = tan2 π₯/2 ππ₯ Putting tan2 π₯/2 = sec2 π₯/2 β 1 π π = ("sec2 " π/π " β 1" )π π Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x β 1 Putting x = π₯/2 tan2 π₯/2 = sec2 π₯/2 β 1 β«1βπ π = β«1β(ππππ π/πβπ) π π y = β«1βγπ ππ2 π₯/2 ππ₯β β«1βππ₯γ y = π/(π/π) tan π/π β x + C y = π tan π/π β x + C is the general solution (As β«1βγπ ππ2 (ππ₯) ππ₯γ = 1/π tan ax + C)

Ex 9.4

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Ex 9.4, 23 (MCQ)

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.